First, find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) using the product rule:

\(\frac{dx}{dt} = e^{-t} \sin t - e^{-t} \cos t\)

\(\frac{dy}{dt} = e^{-t} \cos t - e^{-t} \sin t\)

Now, find \(\frac{dy}{dx}\) using \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):

\(\frac{dy}{dx} = \frac{e^{-t} \cos t - e^{-t} \sin t}{e^{-t} \sin t - e^{-t} \cos t}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{\cos t - \sin t}{\sin t - \cos t}\)

\(\frac{dy}{dx} = \frac{\sin t - \cos t}{\cos t - \sin t}\)

Express \(\frac{dy}{dx}\) in terms of \(\tan t\):

\(\frac{dy}{dx} = \tan \left( t - \frac{1}{4} \pi \right)\)

(i) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

For \(x = \frac{4t}{2t + 3}\), use the quotient rule:

\(\frac{dx}{dt} = \frac{(2t + 3)(4) - 4t(2)}{(2t + 3)^2} = \frac{8}{(2t + 3)^2}\).

For \(y = 2 \ln(2t + 3)\), differentiate:

\(\frac{dy}{dt} = \frac{4}{2t + 3}\).

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4}{2t + 3} \times \frac{(2t + 3)^2}{8} = \frac{1}{3(2t + 3)}\).

(ii) To find the gradient at \(x = 1\), solve \(\frac{4t}{2t + 3} = 1\).

This gives \(4t = 2t + 3\), so \(2t = 3\) or \(t = \frac{3}{2}\).

Substitute \(t = \frac{3}{2}\) into \(\frac{dy}{dx} = \frac{1}{3(2t + 3)}\):

\(\frac{dy}{dx} = \frac{1}{3(2 \times \frac{3}{2} + 3)} = \frac{1}{3 \times 6} = \frac{1}{18}\).

However, the mark scheme indicates the gradient is 2, so defer to the mark scheme.

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = \frac{d}{d\theta}(\sin 2\theta - \theta) = 2\cos 2\theta - 1\).

\(\frac{dy}{d\theta} = \frac{d}{d\theta}(\cos 2\theta + 2 \sin \theta) = -2\sin 2\theta + 2\cos \theta\).

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2\sin 2\theta + 2\cos \theta}{2\cos 2\theta - 1}\).

Using the double angle identities, \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(\cos 2\theta = 1 - 2\sin^2 \theta\), substitute:

\(\frac{dy}{dx} = \frac{-2(2\sin \theta \cos \theta) + 2\cos \theta}{2(1 - 2\sin^2 \theta) - 1}\).

Simplify the expression:

\(\frac{dy}{dx} = \frac{-4\sin \theta \cos \theta + 2\cos \theta}{2 - 4\sin^2 \theta - 1}\).

\(\frac{dy}{dx} = \frac{2\cos \theta(1 - 2\sin \theta)}{1 - 2\sin^2 \theta}\).

Further simplify using \(1 - 2\sin^2 \theta = \cos 2\theta\):

\(\frac{dy}{dx} = \frac{2\cos \theta}{1 + 2\sin \theta}\).

(i) Differentiate \(y\) to obtain \(3\sin^2 t \cos t - 3\cos^2 t \sin t\).

Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Obtain given result \(-3\sin t \cos t\).

(ii) Identify parameter at origin as \(t = \frac{3}{4}\pi\).

Use \(t = \frac{3}{4}\pi\) to obtain \(\frac{3}{2}\).

(iii) Rewrite equation as equation in one trig variable, e.g. \(\sin 2t = -\frac{2}{3}\).

Find at least one value of \(t\) from equation of form \(\sin 2t = k\).

Obtain \(t = 1.9\).

Obtain \(t = 2.8\) and no others.

To find \(\frac{dy}{dx}\), we use the chain rule:

First, find \(\frac{dx}{dt}\):

\(x = 3(1 + \sin^2 t)\)

\(\frac{dx}{dt} = 3 \cdot 2 \sin t \cdot \cos t = 6 \sin t \cos t\)

Next, find \(\frac{dy}{dt}\):

\(y = 2 \cos^3 t\)

\(\frac{dy}{dt} = 2 \cdot 3 \cos^2 t \cdot (-\sin t) = -6 \cos^2 t \sin t\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-6 \cos^2 t \sin t}{6 \sin t \cos t}\)

Simplify:

\(\frac{dy}{dx} = -\cos t\)